Question

(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific to the information in this problem, is it satisfied? Explain why or why not.

(b). If appropriate, use the data in part (a) to calculate and interpret a 99% confidence interval for the proportion of all days in his life that Parsnip has been fed seeds.

Answer 1

1)

Second assumption is that

According to the Rule of Sample Proportions, if np≥10 then the sampling distributing will be approximately normal.

Here np =22 which is greater than 10 and hence condition ssatisfied.

2)

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   22          
Sample Size,   n =    37          
                  
Sample Proportion ,    p̂ = x/n =    0.595          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0807          
margin of error , E = Z*SE =    2.576   *   0.0807   =   0.2079
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.595   -   0.2079   =   0.3867
Interval Upper Limit = p̂ + E =   0.595   +   0.2079   =   0.8025
                  
99%   confidence interval is (   0.3867   < p <    0.8025   )

THANKS

revert back for doubt

please upvote